Problem: ${\sqrt[3]{3645} = \text{?}}$
Solution: $\sqrt[3]{3645}$ is the number that, when multiplied by itself three times, equals $3645$ First break down $3645$ into its prime factorization and look for factors that appear three times. So the prime factorization of $3645$ is $3\times 3\times 3\times 3\times 3\times 3\times 5$ Notice that we can rearrange the factors like so: $3645 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 = (3\times 3\times 3) \times (3\times 3\times 3) \times 5$ So $\sqrt[3]{3645} = \sqrt[3]{3\times 3\times 3} \times \sqrt[3]{3\times 3\times 3} \times \sqrt[3]{5}$ $\sqrt[3]{3645} = 3\times 3 \times \sqrt[3]{5}$ $\sqrt[3]{3645} = 9 \sqrt[3]{5}$